Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 5 down vote favorite

Show that if $A$ and $I+AB$ are invertible, then $I+BA$ is also invertible with $$(I+BA)^{-1} = A^{-1}(I+AB)^{-1}A$$

share|improve this question
Multiply $I+BA$ on the left with $A^{-1}(I+AB)^{-1}A$ try to simplify it using the distributive property. Then do the same on the right. – Tim Jan 26 at 22:13
Three answers, none from me, and yet mine is the only vote for the question so far. – Michael Hardy Jan 26 at 22:16

3 Answers

up vote 11 down vote accepted

Hint: Just do it. Use the fact that

$$ (I + BA) A^{-1} = A^{-1} + B = A^{-1}(I+AB).$$

to show that

$$ (I + BA) A^{-1} (I+AB)^{-1} A = I.$$

share|improve this answer
Is it also necessary to show $A^{-1} (I+AB)^{-1} A (I + BA)= I$? – mezhang Jan 26 at 22:14
2  
@mezhang For general square matrices $M, N$, if we know that $M N = I$, then $N M = I$. – Calvin Lin Jan 26 at 22:16
@mezhang if a square matrix has a one sided inverse, then it has an inverse. CalvinLin: Your comment is only true for square matrices. – Tim Jan 26 at 22:19
@Tim Could you show me a proof for your statement? I don't want to raise a separate question. Thanks. – mezhang Jan 26 at 22:21
@mezhang Check the lemma in my answer here: math.stackexchange.com/questions/280884/… – Git Gud Jan 26 at 22:22
show 2 more comments
up vote 3 down vote

$$ (I + BA) = A^{-1}(I + AB)A \Rightarrow (I + BA)^{-1} = A^{-1}(I + AB)^{-1}A $$

share|improve this answer
up vote 2 down vote

Since $\,A^{-1}C^{-1}A=(A^{-1}CA)^{-1}\,$ , we get with $\,C=I+AB\,$ :

$$(I+BA)\left(A^{-1}(I+AB)^{-1}A\right)=(I+BA)(A^{-1}(I+AB)A)^{-1}=$$

$$(I+BA)(I+BA)^{-1}=I$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.